I am reading Practical Electronics For Inventors 4th edition and on page 384 there is a figure that shows basic power supply with single diode,I think its called half bridge.

There is picture that shows transformer that steps down wall plug ac down to 10 volts.There is written "10 volt AC rms",then it shows single diode,a 1 milliFarad capacitor and its all finally connected to 100 ohm resistor that represents load.

Its kind of a thing where author of the book ask you to calculate and find average dc output. The author says the correct answer is 12.92 V,I absolutely dont understand why its so high.

I calculate voltage like this, average voltage of sine wave is the rms, a 10 V rms AC average voltage is same like 10 V dc average voltage. The picture clearly shows the bottom half of the sine wave completely blocked by diode, only positive peaks are there,now if the sinewave was 10 V rms before it got rectified,then after half of it is chopped away,it must be 5 V rms or 50%! Yet instead of decrease by half the author says its 12.92 or almost 13 V.

With 10 V rms AC, you completely block half of it and then you suddenly have 12.92 V dc?

My question is: Where does the extra voltage come from?

  • \$\begingroup\$ Is it like this? The capacitor helps to keep the voltage up. \$\endgroup\$
    – Octopus
    Mar 22 '17 at 4:47
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    \$\begingroup\$ A 10Vrms sine wave has 28.3V between pos and neg peaks, each peak is 14.14V. OP missed that part? \$\endgroup\$
    – wbeaty
    Mar 22 '17 at 7:17
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    \$\begingroup\$ Also the AVERAGE value of ANY sine wave measured over a whole number of cycles is clearly zero! You need to be really careful about thinking average when you mean RMS (And in this instance the peak value is actually more important). \$\endgroup\$
    – Dan Mills
    Mar 22 '17 at 10:37

You need to look up RMS value and see how it relates to peak to peak...

10VRMS is 28.28V peak-to-peak. 
Half wave rectify that gets you back to 14.14V
Drop the diode voltage gets you to about 13.3V
Average out the ripple on the cap from the 100Ohm resistor
You end up down @ 12.92V

enter image description here


First, make sure you understand the input voltage waveform: 10 V RMS translates to a sine wave with a peak value of \$10 \text {V} \cdot \sqrt{2} \simeq 14.14 \text{ V}_{Pk}\$ (peak to peak value of 28.28 V).

To put it another way, you could describe this waveform by \$10 * \sqrt(2) \sin(60\cdot 2\pi t)\$. When this waveform is half-wave rectified, you now get a waveform with periodic peaks of 14.14 V (diode forward biased), and half cycles of 0 V (diode reverse biased).

To calculate the average voltage on across the resistor (the voltage across the resistor is not constant), you calculate the RC time constant (the capacitor discharging through the resistor when the diode is reverse biased) and where it intersects with the rising peak (when the diode becomes forward biased and the capacitor begins charging again). After solve the intersection, you can now draw the voltage waveform across the resistor. You will find that it is a (mostly)-sawtooth-like waveform, rising to 14.14 V at it's peak, and dropping to ~12 V at it's minimum. The average of the waveform will be 12.92 V, which you can solve by integration or, in a pinch, with a formula for approximating this situation.

From Sedra & Smith:

Half-wave rectifier waveforms

The "shortcut" given is to make sure that \$CR >> T\$, so \$e^{-T/CR} \simeq 1 - T/CR\$ and therefore the ripple voltage \$V_r \simeq V_p \frac{T}{CR}\$.

Keep in mind, this assumes an ideal diode and ideal voltage source.

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    \$\begingroup\$ Towards the end, you gave the exact equation I'm sure the book was looking for. In this case, \$\left(1-\frac{1}{120 R C}\right)\cdot \sqrt{2}\cdot 10\:\textrm{V}_{\textrm{RMS}} \approx 12.92\$. And yeah, that assumes a perfect diode. \$\endgroup\$
    – jonk
    Mar 22 '17 at 6:57

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